Bessel and Legendre Functions Questions and Answers

Q: The integral \(\frac{1}{2\pi i}\oint\frac{\exp[\frac{z}{2}(t - \frac{1}{t})]}{t^{n+1}}dt\), where n is an integer and the closed contour is a circle enclosing the origin, can be used to :

The integral is the contour-integral form of the Bessel functions of the first kind, \(J_{n}(z)\).The generating function of the Bessel functions is\[ \begin{aligned} &\exp\left[\frac{z}{2}\left(t - \frac{1}{t}\right)\right] \\ &= \sum_{n=-\infty}^{\infty} J_{n}(z)\,t^{n} \end{aligned} \]Dividing by \(t^{n+1}\) and integrating round the origin keeps only the \(1/t\) term, so by the residue theorem the integral equals \(J_{n}(z)\), which is the standard contour-integral representation of \(J_{n}\).The other families have different generating functions: \(e^{2xt - t^{2}}\) for the Hermite polynomials and \((1 - 2xt + t^{2})^{-1/2}\) for the Legendre polynomials, from which the associated Legendre functions are obtained by differentiation.

Q: If \(\delta(1-x) = \sum_{n=0}^{\infty} \frac{2n+1}{2} P_{n}(x)\) where \(P_{n}(x)\) are Legendre polynomials and \(\delta\) is usual Dirac-delta function, then \(\delta(1+x)\) can be expressed in terms of \(P_{n}(x)\) as :

The expansion is \(\sum_{n=0}^{\infty}(-1)^{n}\frac{2n+1}{2}P_{n}(x)\).Replace \(x\) by \(-x\) in the given expansion. Then \(\delta(1-(-x)) = \delta(1+x)\), and the Legendre polynomials have definite parity, \(P_{n}(-x) = (-1)^{n}P_{n}(x)\):\[ \begin{aligned} \delta(1+x) &= \sum_{n=0}^{\infty}\frac{2n+1}{2}P_{n}(-x) \\ &= \sum_{n=0}^{\infty}(-1)^{n}\frac{2n+1}{2}P_{n}(x) \end{aligned} \]The same result follows from the coefficient formula \(c_{n} = \frac{2n+1}{2}\int \delta(1+x)P_{n}(x)\,dx = \frac{2n+1}{2}P_{n}(-1)\). The coefficient \((2n+1)/2\) itself does not change, so the options with \(-2n+1\) or \(2n-1\) are wrong, and the option without \((-1)^{n}\) ignores the parity.

The integral \(\frac{1}{2\pi i}\oint\frac{\exp[\frac{z}{2}(t - \frac{1}{t})]}{t^{n+1}}dt\), where n is an integer and the closed contour is a circle enclosing the origin, can be used to :

A.generate Hermite polynomials
B.generate Bessel's functions of first kind
C.generate Legendre polynomials
D.generate associated Legendre polynomials

Exam Year(s)

MH SET - 2023, SET Exam - 2023,

Answer: B

The integral is the contour-integral form of the Bessel functions of the first kind, \(J_{n}(z)\).

The generating function of the Bessel functions is

\[ \begin{aligned} &\exp\left[\frac{z}{2}\left(t - \frac{1}{t}\right)\right] \\ &= \sum_{n=-\infty}^{\infty} J_{n}(z)\,t^{n} \end{aligned} \]

Dividing by \(t^{n+1}\) and integrating round the origin keeps only the \(1/t\) term, so by the residue theorem the integral equals \(J_{n}(z)\), which is the standard contour-integral representation of \(J_{n}\).

The other families have different generating functions: \(e^{2xt - t^{2}}\) for the Hermite polynomials and \((1 - 2xt + t^{2})^{-1/2}\) for the Legendre polynomials, from which the associated Legendre functions are obtained by differentiation.

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For the differential equation

\[ \begin{aligned} &x^{2}\frac{d^{2}y}{dx^{2}} + x\frac{dy}{dx} \\ &+ (x^{2} - n^{2})y = 0 \end{aligned} \]

Which of the following statements is true ?

A.\(x = 0\) and the point at infinity are regular singular points
B.\(x = 0\) is an ordinary point, while the point at infinity is a regular singular point.
C.\(x = 0\) is a regular singular point, while the point at infinity is an irregular singular point
D.\(x = 0\) is an ordinary point, while the point at infinity is singular.

Exam Year(s)

MH SET - 2024, SET Exam - 2024,

Answer: C

\(x = 0\) is a regular singular point, while the point at infinity is an irregular singular point.

This is Bessel's equation. Dividing by \(x^{2}\) gives \(y'' + P(x)y' + Q(x)y = 0\) with \(P = \frac{1}{x}\) and \(Q = 1 - \frac{n^{2}}{x^{2}}\). Both are singular at 0, but \(xP = 1\) and \(x^{2}Q = x^{2} - n^{2}\) are analytic there, so \(x = 0\) is a regular singular point (Frobenius series work).

At infinity, substitute \(x = 1/t\): the transformed equation has a coefficient behaving like \(1/t^{4}\) at \(t = 0\), too strong a singularity to be regular, so infinity is an irregular singular point. That is why Bessel functions decay like \(\frac{\cos x}{\sqrt{x}}\) with an essential oscillation rather than a power law. The options calling \(x = 0\) ordinary are wrong because \(P\) blows up there.

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If \(\delta(1-x) = \sum_{n=0}^{\infty} \frac{2n+1}{2} P_{n}(x)\) where \(P_{n}(x)\) are Legendre polynomials and \(\delta\) is usual Dirac-delta function, then \(\delta(1+x)\) can be expressed in terms of \(P_{n}(x)\) as :

A.\(\sum_{n=0}^{\infty} \frac{-2n+1}{2} P_{n}(x)\)
B.\(\sum_{n=0}^{\infty} (-1)^{n} \frac{-2n+1}{2} P_{n}(x)\)
C.\(\sum_{n=0}^{\infty} (-1)^{n} \frac{2n-1}{2} P_{n}(x)\)
D.\(\sum_{n=0}^{\infty} (-1)^{n} \frac{2n+1}{2} P_{n}(x)\)

Exam Year(s)

MH SET - 2025, SET Exam - 2025,

Answer: D

The expansion is \(\sum_{n=0}^{\infty}(-1)^{n}\frac{2n+1}{2}P_{n}(x)\).

Replace \(x\) by \(-x\) in the given expansion. Then \(\delta(1-(-x)) = \delta(1+x)\), and the Legendre polynomials have definite parity, \(P_{n}(-x) = (-1)^{n}P_{n}(x)\):

\[ \begin{aligned} \delta(1+x) &= \sum_{n=0}^{\infty}\frac{2n+1}{2}P_{n}(-x) \\ &= \sum_{n=0}^{\infty}(-1)^{n}\frac{2n+1}{2}P_{n}(x) \end{aligned} \]

The same result follows from the coefficient formula \(c_{n} = \frac{2n+1}{2}\int \delta(1+x)P_{n}(x)\,dx = \frac{2n+1}{2}P_{n}(-1)\). The coefficient \((2n+1)/2\) itself does not change, so the options with \(-2n+1\) or \(2n-1\) are wrong, and the option without \((-1)^{n}\) ignores the parity.

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